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You may need to use the appropriate appendix table or technology to answer this question
You may need to use the appropriate appendix table or technology to answer this question.
The president of Doerman Distributors, Inc., believes that 25% of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers.
(a)
Assume that the president is correct and p = 0.25.
What is the sampling distribution of p
for n = 100? (Round your answer for σp
to four decimal places.)
σp
=
E(p)
=
Since np =
and n(1 − p) =
, approximating the sampling distribution with a normal distribution ---Select--- is is not
appropriate in this case.
(b)
What is the probability that the sample proportion p
will be between 0.15and 0.35? (Round your answer to four decimal places.)
(c)
What is the probability that the sample proportion will be between 0.20and 0.30? (Round your answer to four decimal places.)
Expert Solution
The mean of the sampling distribution of p is
25% or 0.25
1.)
The standard Deviation of the sampling distribution of p is
(p*(1-p)/n )^1/2
p =0.25
n = 100
= (0.25*(1-0.25)/100)^1/2
= 0.0433
2.)
The Probability that the sample proportion will be between 0.15 and 0.35
= (0.35-0.25)/0.0433
= 2.31
At p = 0.15
= (0.15-0.25)/0.0433
= -2.31
= P(Z <2.31) - P(Z<-2.31)
= by using Excel function
= NORMSDIST(2.31) - NORMSDIST(-2.31)
= 0.9896 - 0.0104
= 0.9791
3.) The Probability that the sample proportion will be between 0.20 and 0.30
= (0.30-0.25)/0.0433
= 1.15
At p = 0.15
= (0.20-0.25)/0.0433
= -1.15
= P(Z <1.15) - P(Z<-1.15)
= by using Excel function
= NORMSDIST(1.15) - NORMSDIST(-1.15)
= 0.8749 - 0.1251
= 0.7499
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